Problem: Find $a+b+c$ if the graph of the equation $y=ax^2+bx+c$ is a parabola with vertex $(5,3)$, vertical axis of symmetry, and contains the point $(2,0)$.
Answer: Since the axis of symmetry is vertical and the vertex is $(5,3)$, the parabola may also be written as  \[y=a(x-5)^2+3\]for some value of $a$.  Plugging the point $(2,0)$ into this equation gives  \[0=a(2-5)^2+3=9a+3.\]This tells us $a=-\frac13$.

Our equation is  \[y=-\frac13(x-5)^2+3.\]Putting it $y=ax^2+bx+c$ form requires expanding the square, so we get  \[y=-\frac13(x^2-10x+25)+3={-\frac13 x^2+\frac{10}{3}x-\frac{16}3}.\]Therefore, $a+b+c = \boxed{-\frac73}$.